The ratio of the forces between two charges placed at a certain distance apart in air and at half of the distance apart in medium of dielectric ‘k’ is

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1 : 4k

4k : 1

k : 4

4 : k

answer is B.

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Detailed Solution

$\frac{F_{1}}{F_{2}} = \frac{\frac{1}{ε_{0}} \frac{1}{r^{2}}}{\frac{1}{{Kε}_{0}} \cdot \frac{1}{(r / 2)^{2}}} \Rightarrow \frac{F_{1}}{F_{2}} = \frac{K}{4}$

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